Quoting Post Archived Topic: It will not be bumped to the top of the forum. Topic factor this fully Membername   Not a member? Sign Up Free (takes 20 seconds) Password   Forgotten your password? Post Font: Verdana Arial Black Georgia Comic Sans Impact   Size: Tiny Normal Big Huge   Color: Default Dark Red Red Orange Brown Green Olive Cyan Blue Dark Blue Indigo Black [ Check Spelling ] [ Post Preview ] [ LiveTags ] Quote: from [m]BleedingSteelWings[/m] at 5:15 pm on July 6, 2009[quote]b)(x^2+16)(x+4) c) (4/25x^2 - 16y^4)(2/5x - 4y^2) I think.[/quote] FAQ Keyword Search: Post Options Favorites Manager Notify me of new replies to this topic by email Notify me of new replies to this topic by private message Original Post youngage Posted at 5:11 pm on July 6, 2009 a) 8x^3 - 1 b) x^3 + 64 c) 8/125x^3 - 64y^6 thanks again

Replies blitzerdog Posted at 8:41 pm on July 6, 2009 No the second and third ones are formulas. (a+b)(a^2-ab+b^2) (a-b)(a^2+ab+b^2) b)(x+4)(x^2-4x+16) c) (2/5x-4y^2)(4/25x^2+8/5xy+16y^4) BleedingSteelWings Posted at 5:38 pm on July 6, 2009 Quote: from youngage at 9:57 pm on July 6, 2009 how do you get it? bleeding steel wings. what do you mean how do you get it? you just take one of the factors out of it. thats how you do the cubed ones. LiveForeverDieTonite Posted at 5:33 pm on July 6, 2009 a) (x+4)(x^2-4x+16) b) (2x-1)(2x^2-2x+1) youngage Posted at 5:27 pm on July 6, 2009 how do you get it? bleeding steel wings. BleedingSteelWings Posted at 5:15 pm on July 6, 2009 b)(x^2+16)(x+4) c) (4/25x^2 - 16y^4)(2/5x - 4y^2) I think. Descartes Posted at 5:15 pm on July 6, 2009 2x+1 2x+1 2x-1 x+4 x+4 x+4 I dont want to do the last one THe first two may be wrong. All 6 previous replies displayed.